Solve x^2-x-30

$${ x }^{ 2 } -x-30$$

Factor

$$\left(x-6\right)\left(x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-30. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-30\right)=-30$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -1.

$$a=-6 b=5$$

Rewrite x^{2}-x-30 as \left(x^{2}-6x\right)+\left(5x-30\right).

$$\left(x^{2}-6x\right)+\left(5x-30\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(x-6\right)+5\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(x+5\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$x^{2}-x-30=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-30\right)}}{2}$$

Multiply -4 times -30.

$$x=\frac{-\left(-1\right)±\sqrt{1+120}}{2}$$

Add 1 to 120.

$$x=\frac{-\left(-1\right)±\sqrt{121}}{2}$$

Take the square root of 121.

$$x=\frac{-\left(-1\right)±11}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±11}{2}$$

Now solve the equation x=\frac{1±11}{2} when ± is plus. Add 1 to 11.

$$x=\frac{12}{2}$$

Divide 12 by 2.

$$x=6$$

Now solve the equation x=\frac{1±11}{2} when ± is minus. Subtract 11 from 1.

$$x=-\frac{10}{2}$$

Divide -10 by 2.

$$x=-5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 6 for x_{1} and -5 for x_{2}.

$$x^{2}-x-30=\left(x-6\right)\left(x-\left(-5\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$x^{2}-x-30=\left(x-6\right)\left(x+5\right)$$

Evaluate

$$\left(x-6\right)\left(x+5\right)$$